A particle of mass 400g is moving under the action of two forces \(F_{1} = (35N, 210°), F_{2} = (35\sqrt{3} N, 300°)\) and a resistance of 40N. Find the mag...
Question 1 Report
A particle of mass 400g is moving under the action of two forces \(F_{1} = (35N, 210°), F_{2} = (35\sqrt{3} N, 300°)\) and a resistance of 40N. Find the magnitude of the
(a) resultant of \(F_{1}\) and \(F_{2}\).
(b) resultant force acting on the particle.
Answer Details
Resolve each force into components \((F\cos\theta,\ F\sin\theta)\).
\(F_{1}=(35,210^{\circ})\): \(\cos210^{\circ}=-\frac{\sqrt3}{2},\ \sin210^{\circ}=-\frac12\), so \(F_{1}=\left(-\tfrac{35\sqrt3}{2},\,-\tfrac{35}{2}\right)\approx(-30.31,\,-17.5)\).
\(F_{2}=(35\sqrt3,300^{\circ})\): \(\cos300^{\circ}=\frac12,\ \sin300^{\circ}=-\frac{\sqrt3}{2}\), so \(F_{2}=\left(\tfrac{35\sqrt3}{2},\,-\tfrac{35\cdot3}{2}\right)\approx(30.31,\,-52.5)\).
(a) Resultant of \(F_{1}\) and \(F_{2}\):
\[R=\big(-30.31+30.31,\ -17.5-52.5\big)=(0,\,-70)\]
Magnitude \(=\sqrt{0^{2}+(-70)^{2}}=70\text{N}\), directed vertically downward (along \(270^{\circ}\)).
(b) The 40N resistance opposes the motion, i.e. it acts opposite to this 70N resultant. The net (resultant) force acting on the particle is therefore
\[70-40=30\text{N}\]
directed the same way as the \(70\)N resultant (\(270^{\circ}\)). Its magnitude is \(\boxed{30\text{N}}\).
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